Question

Va rog o rezolvare la asta? pls (dau coroana)

Answer 1

1. Date:

$m_{s_{HCl}} = 400g$

$c_{HCl} = 35\%$

$n_{H_2}=?$

$A_H=1$

$A_{Cl}=35,5$

Rezolvare:

$c=\dfrac {m_d \cdot 100}{m_s} \implies m_d = \dfrac{c \cdot \:m_s}{100}$

$m_{d_{HCl}} =\dfrac {c_{HCl} \cdot m_{s_{HCl}}} {100} = \dfrac{35 \cdot 4\not0 \not0g}{1 \not0 \not0} = \boxed{140g \: HCl}$

$\mathcal M_{HCl}= A_{H} + A_{Cl} = 1 + 35,5 = 36,5 \dfrac{g}{mol}$

$\mathcal M_{H_2} = 2 \cdot A_{H} = 2 \cdot 1= 2 \dfrac{g}{mol}$

$\\ Zn+ \dfrac { 140g\over{2HCl}} { \small2 \cdot36,5g}→ZnCl_2+\dfrac { x\over { H_2}}{2g}$

$\dfrac{140}{2 \cdot36,5} = \dfrac{x}{2} \implies x = \dfrac{140 \: \cdot \not2}{36,5 \: \cdot\not2} = \boxed {3,83g \: H_2}$

$n=\dfrac {m} {\mathcal M} \implies n_{H_2}=\dfrac {m_{H_2}} {\mathcal M_{H_2}} = \frac{3,83 \not g}{2 \frac{ \not g}{mol}} = \boxed{ \boxed{ 1,91 \: mol \:H_2}}$

2. Vezi atașarea.